STAT 516 hw 5

Author

Karl Gregory

Students in a class were randomly assigned to receive a piece of paper on which was printed a pie chart, a bar chart, or a stacked bar chart (these are shown below). Students were given five seconds to guess, from looking at the chart, what percentage the chart represented for category “B”.

Here are the guesses:

pie,bar,stacked
15.00,10.00,10.00
20.00,17.00,10.00
14.00,13.00,10.00
15.00,25.00,18.00
15.00,25.00,20.00
14.00,25.00,18.00
15.00,16.67,15.00

It is of interest whether there is any difference in the abilities of these types of charts to effectively convey percentages made up by the parts of a whole. The true percentage associated with the category “B” was $15 %$ . Let $Y_{i j}$ be the absolute deviation from $15 %$ of the guess of student $i$ receiving chart type $j$ , and consider the model that treats these responses as independent random variables normally distributed around their respective means.

1.

Import (however it seems best) the data into R and make side-by-side boxplots of the response values (the responses are not the raw guesses but the absolute deviations of the guesses from $15$ ) for the three chart types.

trt <- as.factor(c(rep("pie",7),rep("bar",7),rep("stacked",7)))
pct <- c(15,20,14,15,15,14,15,
         10,17,13,25,25,25,16.37,
         10,10,10,18,20,18,15)
Y <- abs(pct - 15) # absolute difference from 15

boxplot(Y~trt)

2.

Give the mean and standard deviation of the response values as well as the number of replications for each chart type.

means <- aggregate(Y, by = list(trt), mean)
sds <- aggregate(Y, by = list(trt), sd)
ns <- aggregate(Y, by = list(trt), length)

tab <- cbind(means,sds[,2],ns[,2])
colnames(tab) <- c("chart","mean","std. dev.","n")
tab

    chart     mean std. dev. n
1     bar 5.767143  4.124264 7
2     pie 1.000000  1.825742 7
3 stacked 3.714286  1.889822 7

3.

Compute all quantities in the analysis of variance (ANOVA) table below without using the lm() function.

Source	Df	SS	MS	F value	p-value
Treatment	$a - 1$	${SS}_{Trt}$	${MS}_{Trt}$	$F_{test}$	$P (F > F_{test})$
Error	$N - a$	${SS}_{Error}$	${MS}_{Error}$
Total	$N - 1$	${SS}_{Tot}$

Y.. <- mean(Y)
Yi. <- aggregate(Y ~ trt, FUN=mean)[,2]
n <- aggregate(Y ~ trt,FUN = length)[,2]
a <- length(Yi.)
N <- sum(n)

SStot <- sum((Y - Y..)**2)
SStrt <- sum(n*(Yi. - Y..)^2)
SSerror <- SStot - SStrt

MStrt <- SStrt / (a - 1)
MSerror <- SSerror / (N - a)

Ftest <- MStrt / MSerror
pval <- 1 - pf(Ftest, a-1, N - a)

4.

Obtain the ANOVA table using the lm() and anova() function (from here on you may use the lm() function).

lm_out <- lm(Y ~ trt)
anova(lm_out)

Analysis of Variance Table

Response: Y
          Df Sum Sq Mean Sq F value Pr(>F)  
trt        2  80.05  40.025  5.0211 0.0185 *
Residuals 18 143.49   7.971                 
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

5.

Give the null hypothesis for which the value $F_{test}$ in the ANOVA table is the test statistic.

The null hypothesis is that all the chart type means are the same.

6.

State your conclusions with respect to the null hypothesis in the previous part.

We will reject the null hypotheses at 
any significance level greater than 0.01849927, 
so the evidence is quite strong that there is 
some difference among the means for the 
different chart types.

7.

Check whether the assumptions of the one-way ANOVA model are satisfied.

plot(lm_out,which=1, add.smooth = F)

plot(lm_out,which = 2)

The assumption of equal variances is 
dubious.  The assumption of normality is also 
somewhat dubious, as the normal quantile-quantile plot 
exhibits some deviations from normality in the 
residuals.

8.

Conduct Levene’s test for equal variances in the response values across chart types.

library(car)

Warning: package 'car' was built under R version 4.4.3

Loading required package: carData

leveneTest(Y ~ trt, center = mean)

Levene's Test for Homogeneity of Variance (center = mean)
      Df F value   Pr(>F)   
group  2  8.4852 0.002536 **
      18                    
---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Levene's test rejects the null hypothesis 
of equal variances.

9.

Give the values of $\hat{μ}$ and of the ${\hat{τ}}_{i}$ in the estimated model and interpret their values.

lm_out


Call:
lm(formula = Y ~ trt)

Coefficients:
(Intercept)       trtpie   trtstacked  
      5.767       -4.767       -2.053

The estimate for mu, 5.767143, is the 
mean of the bar chart responses, the estimate for 
tau_2, -4.767143, is the mean of the pie 
chart responses minus the mean of the bar chart 
responses, and the estimate for 
tau_3, -2.052857, is the mean of the stacked bar chart 
responses minues the mean of the bar chart 
responses.

10.

Proceeding as though the assumptions were satisfied, perform a careful analysis in order to determine, if possible, which of the three chart types is best for giving an accurate impression of a percentage. Give a careful summary of your findings.

# do tukey's
aov_out <- aov(lm_out)
tukey_out <- TukeyHSD(aov_out)
tukey_out

  Tukey multiple comparisons of means
    95% family-wise confidence level

Fit: aov(formula = lm_out)

$trt
                 diff       lwr        upr     p adj
pie-bar     -4.767143 -8.618757 -0.9155285 0.0142951
stacked-bar -2.052857 -5.904471  1.7987572 0.3817976
stacked-pie  2.714286 -1.137329  6.5659000 0.1982937

plot(tukey_out)

To answer this question we should 
compare all pairs of means with Tukey's method. 
The result of this shows that the pie chart 
is better than the bar chart; however, the 
pie chart cannot be said to be better than 
the stacked bar chart, nor is the difference 
between the bar and the stacked bar charts statistically significant. We may 
say tentatively that the pie chart emerges 
as a possible winner; but one would have to 
collect more data in order to make this claim 
with great confidence.

11.

Now take the pie chart as a benchmark against which the other two chart types are to be compared. Investigate carefully whether the ability of the pie chart to convey a percentage is different from that of the bar and the stacked bar charts. Carefully summarize your findings.

library(DescTools)

Warning: package 'DescTools' was built under R version 4.4.1


Attaching package: 'DescTools'

The following object is masked from 'package:car':

    Recode

DunnettTest(Y ~ trt, control="pie")


  Dunnett's test for comparing several treatments with a control :  
    95% family-wise confidence level

$pie
                diff     lwr.ci   upr.ci   pval    
bar-pie     4.767143  1.1467997 8.387486 0.0102 *  
stacked-pie 2.714286 -0.9060574 6.334629 0.1547    

---
Signif. codes:  0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1

Now we compare the bar and stacked 
bar chart means with the pie chart mean 
using Dunnett's method.  The results are 
scarely different from those of Tukey's method 
for comparing all pairs of means.

12.

Should we have expected the findings from parts 10. and 11. to be very different? Why or why not?

We should not expect the findings 
to be very different, since, given that there 
are only three treatment groups, Tukey's 
method makes 3 comparisons and Dunnett's makes 
two comparisons.  Since the numbers of 
comparisons made by the two methods differ only by 
one, the methods yield quite similar 
results.  Dunnett's will give confidence 
intervals narrower than the Tukey intervals by a 
greater margin if the number of treatment 
groups is larger.